# converting integrand limits from Cartesian to spherical

I am trying to find a general formula for volume of a ball given by $$\{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 < a^2 \}$$

The triple integral I am trying to solve is $$\iiint_B{(x^2+y^2+z^2)\mathrm dV}$$

So from what I know I have to convert this to spherical to make an easy iterated integral. I know $p^2 = x^2 +y^2 +z^2$ and so our limits have to go from $0 < p < a, 0 < \theta<2\pi \text{ and } 0 < \phi < \pi$

So, am I solving the following iterated integral? $$\int_0^{2\pi}\int_0^\pi \int_0^a{p^2 \cdot p^2\mathrm dp\mathrm d\phi\mathrm d\theta}$$

-
I think the downvoter should explain the reason for the downvote. – Srivatsan Oct 6 '11 at 23:16

My earlier answer consisting only of a "yes" referred to the correct transformation of the integral in the second displayed equation to the one in the third. I see now that you actually started out with a different question, namely to find a general formula for the volume of a ball, and in a comment you added a slightly different focus on an explanation on how to get the limits.

The integral in the second displayed equation is not the volume of the ball defined by the first. The volume element is $\mathrm dV$, so the volume is the integral over $\mathrm dV$, not over $r^2\mathrm dV$. (I've never seen the variable $p$ used for the radial coordinate before; I'll use $r$.) The expression for $r^2$ appears in the inequality bounding the ball; there's no reason to put it into the integrand.

As to a general approach to finding the limits: You need to reexpress the inequalities bounding the integration region in the new coordinates. In the present case, this is particularly simple, since the left-hand side of the only inequality is just $r^2$, so it becomes $r^2\lt a^2$, and there are no further bounds, so the angular variables take their full range.

You can see a more involved example where all coordinates occur in the inequalities bounding the volume at Volume of Region in 5D Space.

Regarding your result $\frac25\pi^2a^2$, this can't be the volume of the ball because that would have to vary with $a^3$, and it can't be the value of the integral you wrote because that would have to vary with $a^5$.

-
I am mainly looking for an explanation on how to get the limits. Geometrically maybe? Why over 2 Pi and 0 to pi – Tyler Hilton Oct 7 '11 at 1:41
Also I got $\frac{2}{5}\cdot \pi^2 \cdot a^2$ – Tyler Hilton Oct 7 '11 at 5:01
In transfirormation do I have to add a sin factor? – Tyler Hilton Oct 7 '11 at 14:40
@Tyler: Yes, I'm sorry, I overlooked that. There's a $\sin \phi$ missing in the integrand, which comes from the Jacobian determinant (as does the factor $r^2$). – joriki Oct 7 '11 at 15:05